mysql - 将数据库中的数据填充到选择的保管箱中

Question

我有一个需要从数据库中获取的选择框。

查看 Blade :

<select class="form-control" name="service_type" id="service_type" data-parsley-required="true">
@foreach ($service as $sn) 

{
 <option value="">dd($sn)</option>
}
@endforeach
</select>

Controller :

public function index()
{
    $service = DB::select('select service_type from services');

    return view("service",['service'=>$service]);
}

我收到类似的错误

Undefined variable: service (View: /home/devadmin/.config/composer/vendor/laravel/installer/serenseprj/resources/views/package.blade.php)

Answer 1

您可以使用 Eloquent 来代替这样做

$service = DB::select('select service_type from services');

使用 Eloquent ，

$services = ModelName::pluck('name', 'id')->get();

您可以将变量传递到目标 View ，如下所示:

return view('view_name', ['variable_name' => $variable]);

并且，您可以像下面这样访问变量名称

<select class="form-control" name="service_type" id="service_type" data-parsley-required="true">
@foreach ($variable_name as $var)
{
 <option value="{{$var->id}}">{{$var->name}}</option>
}
@endforeach
</select>